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gopeder

  • 3 years ago

Hey guys I have a few problems related to a figure I'd like to understand the problem and learn how to solve it, I think Im just a little unclear on it but anyways any help is welcomed :)

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  1. gopeder
    • 3 years ago
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  2. gopeder
    • 3 years ago
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    Im trying it out again i think i may have just mistaken something

  3. gopeder
    • 3 years ago
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    1 is choice 3: 112

  4. gopeder
    • 3 years ago
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    Ok 2 3 & 4 lost me :S

  5. saifoo.khan
    • 3 years ago
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    how can u say 1) 112?

  6. gopeder
    • 3 years ago
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    can someone help break them down a bit plz?

  7. saifoo.khan
    • 3 years ago
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    i dont think 1) = 112..

  8. saifoo.khan
    • 3 years ago
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    180 - 112 = ?

  9. gopeder
    • 3 years ago
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    well ACF covers the same length as GAC

  10. saifoo.khan
    • 3 years ago
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    Nope, |dw:1331792427015:dw|

  11. saifoo.khan
    • 3 years ago
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    the lengths dosnt really matter..

  12. gopeder
    • 3 years ago
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    oh its the angels not the lengths of the line :\

  13. gopeder
    • 3 years ago
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    ps sorry open study kinda crashed google chrome

  14. gopeder
    • 3 years ago
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    So how do i go about finding the angle lengths for the first problem:? so I can figure out Part one of the question?

  15. AccessDenied
    • 3 years ago
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    its a parallelogram, so AG is parallel to CF with a transversal AC, and m< GAC = 112 |dw:1331792928009:dw|

  16. AccessDenied
    • 3 years ago
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    soo that would make ACF the "same-side interior angle" to GAC, which means they will be supplementary. That makes sense?

  17. gopeder
    • 3 years ago
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    so i was right m<ACF = 112 because ACF is parallel to GAC

  18. S
    • 3 years ago
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    1. |dw:1331793329783:dw| Do you see how GAC and CFG look the same? Meaning they are both more then 90 degrees.. they are opposite or something i forgot what its called cause its been a while but they are opposite and equal to each other. Same way ACF and AGF are equal to each other.. they are both sharp.. both less then 90 degrees.. Now the sum of all 4 angles is 360. AGF + GAC = 180 or ACF + CFG = 180 or CFG + FGA = 180 or GAC + ACG = 180

  19. S
    • 3 years ago
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    As we said: GAC + ACF = 180 They gave us that GAC = 112, So lets plug in and solve for ACF 112 + ACF = 180 ACF = 180 - 112 ACF = 68

  20. gopeder
    • 3 years ago
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    Wow that was vary clear Id like help on the other problems, but first can you test me on a another problem like this one? i want to see if i really got it

  21. S
    • 3 years ago
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    Ok I ll help you with the rest of them. Yes, lets try a similar problem. |dw:1331794002046:dw| Angle D is equal to 73 degrees. Find: Angle A Angle B Angle C

  22. gopeder
    • 3 years ago
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    B has to be 73 and A & C must be 107 theya ll then add up to 360

  23. S
    • 3 years ago
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    Yup! That's perfect!

  24. gopeder
    • 3 years ago
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    YEA!!! :)

  25. S
    • 3 years ago
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    So all of them add up to 360 or like neighboring pairs add up to 180.. like A+B = B+C = C+D = D+A = 180

  26. S
    • 3 years ago
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    Ok lets do number 2 now =)

  27. S
    • 3 years ago
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    We are dealing with AGF and GAC. Just by looking at them we know that they add up to180.

  28. S
    • 3 years ago
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    So if AGF + GAC = 180, lets now plug in the values that are given to us (a+20) + (2a+10) = 180 a + 20 + 2a + 10 = 180 3a + 30 = 180 3a = 180 - 30 3a = 150 a = 50 Now we need to find GFC which is the same as GAC which they said is (2a + 10) Plug in 50 for a 2a+10 = 2*50 + 10 = 110 degrees

  29. S
    • 3 years ago
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    So our GFC, (as well as GAC) is equal to 110 degrees

  30. S
    • 3 years ago
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    For number 3 |dw:1331794863730:dw| AC and GF have to equal to each other. In a parallelogram, opposite sides are parallel and equal, so AC = GF and AG = CF

  31. S
    • 3 years ago
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    AG = CF 2x + 10 = 5x - 2 10 + 2 = 5x - 2x 12 = 3x x = 12/3 x = 4

  32. gopeder
    • 3 years ago
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    Ok I get 3 but not 2 lol 2 was a mind Puck

  33. S
    • 3 years ago
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    Ok question 2 Look at angles AGF and GAC. They are neighbors.. one looks like he is less then 90, other one looks to be more then 90 correct? So they add up to 180. Do you understand this part?

  34. gopeder
    • 3 years ago
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    Yes I do

  35. S
    • 3 years ago
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    Ok, so AGF = GAC In the question they said that AGF = (a+20) degrees and GAC = (2a+10) degrees correct?

  36. gopeder
    • 3 years ago
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    a+20 <= this has me saying wtf

  37. gopeder
    • 3 years ago
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    is it just there to add to the problem?

  38. S
    • 3 years ago
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    Yea I know, but just look at it as a regular number. You know like you have for example x + 2 = 3 x is just a hidden unknown number.. and you find it by x = 3 - 2... x = 1 Same thing here.. instead of giving you a regular number, they give a hidden unknown number, which in this case is not (x), but an (x + 20), or as they put it an (a + 20)

  39. S
    • 3 years ago
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    So instead of giving you AGF = 70 degrees for example, they give you AGF = (a+20) degrees. They want you to find a and then use it in order to find the angle

  40. S
    • 3 years ago
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    And the only way we can solve for from this is to set it equal to something that we know it actually equals. In this case, sum of these two angles AGF and GAC is equal to 180

  41. gopeder
    • 3 years ago
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    OK err thats still going to get me its just one of those things lol but i understand it now so thats good:)

  42. S
    • 3 years ago
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    |dw:1331795961782:dw|

  43. S
    • 3 years ago
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    Ok, well its just the same as we had before. One angle plus another is equal to 180. In this case instead of having numerical values we have unknown and numerical values, but by putting them together its still equal to 180. So from there we find our unknown a: So if AGF + GAC = 180, lets now plug in the values that are given to us (a+20) + (2a+10) = 180 a + 20 + 2a + 10 = 180 3a + 30 = 180 3a = 180 - 30 3a = 150 a = 50 Now we need to find GFC which is the same as GAC which they said is (2a + 10) Plug in 50 for a 2a+10 = 2*50 + 10 = 110 degrees

  44. S
    • 3 years ago
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    Ok where we left off? 3 or 4? =D

  45. gopeder
    • 3 years ago
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    4 And man your My hero :)

  46. S
    • 3 years ago
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    Hehe, no problem man, don't mention it =D This stuff helps your brain, and exercises mine =D So I benefit from it as much as you do =D

  47. S
    • 3 years ago
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    Ok now question 4. |dw:1331796416239:dw| First look at the square ABCD. Its diagonals AC and BD bisect each other. They intersect at point O and cut each other in half. So AO = CO = BO = DO

  48. S
    • 3 years ago
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    Now it works similarly with a parallelogram ACGF. AF and CG intersect each other at point X. They bisect each other, or cut each other in half. So, AX = XF and CX = XG. Note that this part is not the same as the one we had with a square, where aaaallllll the little pieces AO BO CO DO were equal to each other. Here, only AX = XF and CX = XG, because if you see, in a parallelogram, diagonals are not equal, AF is shorter than CG. So once again, here AX = XF and CX = XG.

  49. S
    • 3 years ago
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    Now lets get to our question. We know that AX = XF Now we plug in the given values for AX and XF (4y+3) = (2y+7) And from here we solve for y. 4y + 3 = 2y + 7 4y - 2y = 7 - 3 2y = 4 y = 2

  50. gopeder
    • 3 years ago
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    First of all Id like to sum up my thoughts on this, Loads of steps lol looks complex at first sight but it is simple when someone smart breaks it down :) I would also like to give a BIG Thank you And let you know I have learned a lot in the pas hour working with you

  51. gopeder
    • 3 years ago
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    Really I cant thank you enough :) I really got it & my friend it was all thanks to you :-D

  52. S
    • 3 years ago
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    Hehe, yea, it may seem confusing at first, but that's all it is about, breaking it down into small and simple parts and solving it nicely and simply =) Noo problem man, don't worry about it, I'm glad I could help you =) That's why we are on this website in the first place =D Give help and get help =D

  53. gopeder
    • 3 years ago
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    Amen to that man :)

  54. S
    • 3 years ago
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    Lol =D

  55. S
    • 3 years ago
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    Ok, anytime you need help with this type of stuff, geometry and stuff, let me know, send me a message or something, I ll try my best to explain it to you )

  56. gopeder
    • 3 years ago
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    Again id like to let you know im walking away with a better understanding of how to better dissect problems :) And i will :) ps you should add me as a fan xD so i can see when your on

  57. S
    • 3 years ago
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    Ok, glad I could help you =) Yup, good idea, just added you! =)

  58. S
    • 3 years ago
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    Ok, I'm gonna take a look at a few more problems and go to sleep now =D

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